Deserialize optional polymorphic fields with jackson

Question

Problem

For example, I have a list of some products in my cart like cinema ticket, carsharing and new book.

[
  {
    "name": "Cinema Ticket"
  }, 
  {
    "name": "Car Sharing",
    "properties": { ... }
  },
  {
    "name": "New Book",
  }
]

As you can see, not all products have properties . Note: this field is polymorphic.

Question

Can I turn "non-existing" field properties into null using Jackson or it's better to change api? And if I can then how to do that?

Jackson Version: 2.10.1

Thanks for answers!

Answer 1

I suppose you have something like:

public class BaseProduct {
   public String name;
}

public class CarSharing extends BaseProduct {
    public String properties;
}

public class Book extends BaseProduct {
}

Instead of having the properties field set to null when is not there, you can use Jackson polymorphic functionality to not show it. Something like:

@JsonTypeInfo(use = JsonTypeInfo.Id.MINIMAL_CLASS, include = JsonTypeInfo.As.PROPERTY, property = "type")
public class BaseProduct {
     public String name;
}

The output would be:

[
  {
    "type": ".Book",
    "name": "Book name"
  },
  {
    "type": ".CarSharing",
    "name": "Car share name",
    "properties": "a property field"
  }
]